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I am trying to prove that, for the real scalar field $\phi(x)$, the retarded propagator, which is defined as $$ D_{R}(x-y)=\theta(x^0-y^0)\langle 0 |[\phi(x),\phi(y)]|0\rangle $$ is the Green's function for the Klein-Gordon operator; that is the function $G_R(x-y)$ such that $$ (\partial_{\mu}\partial^{\mu}-m^2)G_F(x-y)=i\delta^{(4)}(x-y). $$ Here the metric is $\eta_{\mu\nu}=\text{diag}(-,+,+,+)$. I am proceeding as follows: $$ (\partial_{\mu}\partial^{\mu}-m^2)D_{R}(x-y)=(\partial_{\mu}\partial^{\mu}\theta(x^0-y^0))\langle 0 |[\phi(x),\phi(y)]|0\rangle+\theta(x^0-y^0)[(\partial_{\mu}\partial^{\mu}-m^2)\langle 0 |[\phi(x),\phi(y)]|0\rangle]+2\partial_{\mu}\theta(x^0-y^0)\partial^{\mu}\langle 0 |[\phi(x),\phi(y)]|0\rangle. $$ The first term is zero because we get $$ \partial_{\mu}\partial^{\mu}\theta(x^0-y^0))\langle 0 |[\phi(x),\phi(y)]|0\rangle=-\partial_{0}(\delta(x^0-y^0))\langle0 |[\phi(\vec{x},x^0),\phi(\vec{y},x^0)]|0\rangle=0. $$ The second term is also zero because it satisfies KG's equation. For the last term we can write $$ 2\partial_{\mu}\theta(x^0-y^0)\partial^{\mu}\langle 0 |[\phi(x),\phi(y)]|0\rangle=-2\delta(x^0-y^0)\langle 0 |[\pi(\vec{x},x^0),\phi(\vec{y},x^0)]|0\rangle=2i\delta(x^0-y^0)\delta^{(3)}(\vec{x}-\vec{y})=2i\delta^{(4)}(x-y). $$ From this I see that $G_R(x-y)=\frac{D_{R}(x-y)}{2}$. There is a factor of two which makes me think I'm forgetting something, either this or Green's function isn't exactly the propagator but only proportional to it. I think this is relevant because the same thing happens for Feynman propagator and Feynman Green's function, which are linked to scattering amplitudes trough the LSZ formula.

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2 Answers 2

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If ${\scr D}$ is some differential operator, a Green's function is a bidistribution $G(x,y)$ with the property that it solves the equation $${\scr D}_xG(x,y)=\delta(x,y)\tag{1}.$$

That is a definition. Now the thing is that to uniquely define $G(x,y)$ one must prescribe extra conditions. In particular, when ${\scr D}=\Box-m^2$ is the Klein-Gordon operator, the standard $i\varepsilon$ prescription is one such condition when solving (1) which has the property that the obtained solution is the time-ordered two-point function $\langle 0|T\{\phi(x)\phi(y)\}|0\rangle$ in free theory. That would be the Feynman propagator. The retarded/advanced propagators follow from other conditions added to (1).

Some authors may define $G(x,y)$ by including some numeric factor to the RHS of (1), but then one is changing the definition, and one must be consistent the whole way through.

So the short answer is that the retarded propagator is one possible Green's function, as is the Feynman propagator. They are all solutions to (1) which differ by some subsidiary condition which picks one out of the space of inverses of ${\mathscr{D}}$.

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Retarded propagator is a Green's function, but a Green's function is not necessarily a propagator. There are two reasons why a green's function might not be a retarded propagator:

  • Boundary conditions in time, which allow introducing retarded, advanced, time-ordered and anti-time-ordered Green's functions (one could also add lesser, greater and Keldysh functions, but only three of the all named are independent).
  • Propagator is usually a one-particle Green's function, whereas we may deal also with many-particle Green's functions.
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